Ammonium iodide dissociates reversibly to ammonia and hydrogen iodide.
NH4I(s) <==> NH3(g) + HI(g)
At 400°C, Kp = 0.215. Calculate the partial pressure of ammonia at equilibrium when a sufficient quantity of ammonium iodide is heated to 400°C.
Please help! would it just be the same Kp since it says its being heated to the same degree?
No, the partial pressure of ammonia at equilibrium will not be the same as the given Kp value because \(K_p\) is a constant that is specific to a particular equilibrium reaction at a given temperature. It is not an indicator of the partial pressures of the gases involved.
To determine the partial pressure of ammonia at equilibrium, we need the initial partial pressure of NH4I and the stoichiometry of the reaction.
Since it is stated that a sufficient quantity of ammonium iodide is heated to 400°C, we can assume that NH4I is the only source of ammonia and hydrogen iodide gases. Therefore, the initial partial pressure of NH4I can be considered equal to the total pressure at equilibrium.
Let's assume that the initial pressure of NH4I is \(P_0\), and \(P_{\text{{NH3}}}\) and \(P_{\text{{HI}}}\) represent the partial pressures of ammonia and hydrogen iodide at equilibrium, respectively.
According to the balanced equation, the stoichiometry is as follows:
1 mol NH4I produces 1 mol NH3 and 1 mol HI.
At equilibrium, the partial pressures of NH3 and HI will be equal. Therefore, let's assume that both are \(P\) at equilibrium.
Using the information above, we can set up the expression for \(K_p\):
\(K_p = \frac{{P_{\text{{NH3}}} \cdot P_{\text{{HI}}}}}{{(P_0 - P_{\text{{NH3}}})}}\)
Since the equation is given as a reversible reaction, we can assume that the forwards and backwards rates of reaction are equal at equilibrium. Therefore, the moles of NH3 produced will be equal to the moles of NH4I consumed.
Given that the initial pressure of NH4I is \(P_0\), and assuming the volume is constant, the pressure of ammonia will be \(P_{\text{{NH3}}} = P - P_0\).
Now, substitute these values into the equation for \(K_p\):
\(0.215 = \frac{{(P - P_0) \cdot (P - P_0)}}{{(P_0 - (P - P_0))}}\)
Simplifying this equation will give you a quadratic equation, which you can solve to find the value of \(P_{\text{{NH3}}}\) at equilibrium.
To calculate the partial pressure of ammonia at equilibrium, we need to use the equilibrium expression and the given value of Kp.
The equilibrium expression for the reaction is: Kp = (P(NH3) * P(HI)) / P(NH4I)
Since the reaction is reversible, the forward and reverse reactions are taking place simultaneously. Therefore, at equilibrium, the partial pressure of each component is related to the stoichiometric coefficients in the balanced equation.
In this case, NH3 and HI have a stoichiometric coefficient of 1, while NH4I has a stoichiometric coefficient of 1 as well. So the equilibrium expression can be simplified to:
Kp = P(NH3) * P(HI) / P(NH4I)
We are given that at 400°C, Kp = 0.215. Since the reaction is being heated to the same degree, we assume that the temperature is at equilibrium, which allows us to use this Kp value.
Now, we know that the partial pressure of NH4I is 1 atm (since it is in a sufficient quantity). Let's assume that the partial pressure of NH3 at equilibrium is P(NH3).
According to the equilibrium expression, if we substitute the given values, we have:
0.215 = P(NH3) * P(HI) / 1
To find the partial pressure of NH3 at equilibrium, we need to solve for P(NH3).
Rearranging the equation, we get:
P(NH3) = 0.215 * 1 / P(HI)
However, we don't have the value of P(HI) given, so we cannot calculate the partial pressure of NH3 without that information.
Therefore, we need additional data (e.g., the value of P(HI) or information about the molar concentrations of NH3, HI, and NH4I) in order to calculate the partial pressure of ammonia at equilibrium.
The problem gives you the Kp at 400 C. We presume the reaction is at the same temperature as Kp is given
Kp = pNH3*pHI
NH4I(s) ==> NH3(g) + HI(g)
..............p.......p
Substitute p into Kp expression and solve for p.